What is the integral from 0 to 2 of 4.0553e^{1.8t} ?

The integral from 0 to 2 of 4.0553e^{1.8t} is 80.20084…

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integral from 0 to 2 of 4.0553e^{1.8t}

\int_{0}^{2} 4.0553 e^{1.8 t} d t

80.20084 \dots

Solution steps

\int_{0}^{2} 4.0553 e^{1.8 t} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4.0553 \cdot \int_{0}^{2} e^{1.8 t} d t

Apply u-substitution

: \int_{0}^{3.6} e^{u} \frac{1}{1.8} d u

= 4.0553 \cdot \int_{0}^{3.6} e^{u} \frac{1}{1.8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4.0553 \cdot \frac{1}{1.8} \cdot \int_{0}^{3.6} e^{u} d u

Simplify

= 4.0553 \cdot 0.55555 \dots \cdot \int_{0}^{3.6} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= 4.0553 \cdot 0.55555 \dots [e^{u}]_{0}^{3.6}

Simplify

= 2.25294 \dots [e^{u}]_{0}^{3.6}

Compute the boundaries:

35.59823 \dots

= 2.25294 \dots \cdot 35.59823 \dots

Simplify

= 80.20084 \dots

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\frac{1}{3 - 2 x}

\frac{d ^{4}}{d x ^{4}} (\frac{1}{x})

What is the integral from 0 to 2 of 4.0553e^{1.8t} ?
The integral from 0 to 2 of 4.0553e^{1.8t} is 80.20084…